An object is placed 20.0m from a converging lens with a focal length 15 cm. A co

Question

An object is placed 20.0m from a converging lens with a focal length 15 cm. A concave mirror with focal length 10cm is located 75cm to the right of the lens as shown in the figure. (Please refer to: http://www.njaapt.org/Physics%20Competitions/Science%20League/Physics1Tests/April/Physics%201%20April%202006.pdf pg. 5 # 23. A) Determine the location of the final image. B) If the height of the object is 1.0cm, what is the height of the image? C) How would you describe the final image (size, orientation)? An object is placed 20.0m from a converging lens with a focal length 15 cm. A concave mirror with focal length 10cm is located 75cm to the right of the lens as shown in the figure. (Please refer to: http://www.njaapt.org/Physics%20Competitions/Science%20League/Physics1Tests/April/Physics%201%20April%202006.pdf pg. 5 # 23. A) Determine the location of the final image. B) If the height of the object is 1.0cm, what is the height of the image? C) How would you describe the final image (size, orientation)?

Explanation / Answer

a.>for a converging lens, f is positive f = 15 u = -20 1/f = 1/u + 1/v =>1/15 = -1/20 + 1/v => 1/v = 1/15 +1/20 = 7/60 =>v = 60/7 m1 = v/u = (60/7)/(-20) = -3/7 for concave mirror, u = -(70-60/7) = -430/7 f = -10 1/f = 1/v -1/u 1/(-10) = 1/v + 7/430 =>1/v = 1/(-10) -7/430 = -50/430 => v = -430/50 = -8.6 cm so, image position is 8.6 cm from concave mirror on its left side. m2 = -v/u = -(-8.6)/(-430/7) = -7/50 final magnification m = m1*m2 = -3/7 * -7/50 = 0.06 c.)so, the final image is upright because magnification is positive and also the image formed is a real image.

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